ReciQues-04-1-Key.pdf_, filename=_ReciQues-04-1-Key.pdf_, filename=_ReciQues-04-1-Key.pdf_, filena

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Chem 109 Recitation Problems – Chapter 4 Part 1 Answer Key 1 1. Solid tetraphosphorus decoxide (P 4 O 10 ) reacts with liquid water to form an aqueous solution of phosphoric acid. P 4 O 10 (s) + 6 H 2 O (l) ! 4H 3 PO 4 (aq) A. How many grams of phosphoric acid will form when 36.00 grams of tetraphosphorus decoxide reacts with 94.00 grams of water? 36.00 g 1 molP 4 O 10 283.88 gP 4 O 10 4 molH 3 PO 4 1 molP 4 O 10 97.994 gH 3 PO 4 1 molH 3 PO 4 = 49.71 gH 3 PO 4 94.00 gH 2 O 1 molH 2 O 18.016 gH 2 O 4 molH 3 PO 4 6 molH 2 O 97.994 gH 3 PO 4 1 molH 3 PO 4 = 340.9 gH 3 PO 4 Since the P 4 O 10 produces less H 3 PO 4 , it is the limiting reagent and the amount of H 3 PO 4 calculated from the P 4 O 10 will be the mass produced. B. What is the limiting reagent in the reaction? Since the P 4 O 10 produces less H 3 PO 4 , it is the limiting reagent C. In this reaction, two reactants form one product. According to the Law of Conservation of Mass, matter cannot be created and/or destroyed during a
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This note was uploaded on 04/12/2008 for the course CHEM 109 taught by Professor Malina during the Spring '08 term at UNL.

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